Question

The equations of motion of a projectile are given by $x = 36t\, m$ and $2y = 96t - 9.8t^2\,m$. The angle of projection is

• A. $sin^{-1}\left(\frac{4}{5}\right)$
• B. $sin^{-1}\left(\frac{3}{5}\right)$
• C. $sin^{-1}\left(\frac{4}{3}\right)$
• D. $sin^{-1}\left(\frac{3}{4}\right)$

Answer 1

Answer: (A)

$sin^{-1}\left(\frac{4}{5}\right)$

Answer 2

Given : $x = 36t$, $2y = 96t - 9.8t^2$ or $y = 48t-4.9t^2$ Let the initial velocity of projectile be $u$ and angle of projection is $\theta$. Then, Initial horizontal component of velocity, $u_{x}=u\,cos\,\theta=\left(\frac{dx}{dt}\right)_{t = 0}=36$ or $ucos\theta = 36\,...\left(i\right)$ Initial vertical component of velocity, $u_{y}=u\,sin\,\theta=\left(\frac{dy}{dt}\right)_{t=0}=48$ or $usin\theta = 48\,... \left(ii\right)$ Dividing $\left(ii\right)$ by $\left(i\right)$, we get $tan\,\theta=\frac{48}{36}=\frac{4}{3}\,$ $\therefore sin\,\theta=\frac{4}{5}$ or $\theta=sin^{-1}\left(\frac{4}{5}\right)$